Chapter 10, Problem 30RE

Solution

a.

To Calculate: The median, range, interquartile range and five-number summary.

The median is $65$ .

The range is $199$ .

The Interquartile Range is $70$ .

The five-number summary is $\{39,45,65,115,238\}$ .

Given information:

The number of visitors per month is:

Visitors
238
211
181
177
176
146
115
115
105
98
79
67
63
62
55
46
46
45
44
44
43
43
40
39
65

Calculation:

The median is calculated as:

Arrange the data in ascending order and the number of observations are $n=17$ .

  $\begin{array}{c}Median(X)=\frac{X_{\frac{n}{2}+1}+X_{\frac{n}{2}}}{2},\text{if }n\text{ is even}\\ \text{=}{X}_{\frac{n+1}{2}},\text{if n is odd}\end{array}$

Here $n=25$ is odd.

  $\begin{array}{c}Median(X)=X_{13}\\ =65\end{array}$

The range is calculated as:

  $\begin{array}{c}Range(X)=Maximum-Minimum\\ =238-39\\ =199\end{array}$

The Interquartile range is calculated as:

  $IQR=Q_3-Q_1,\text{ where }{Q}_1\text{ is median of first of data and }{Q}_3\text{ is median of lower half of the data}\text{.}$

  $\begin{array}{c}{Q}_1=Median(X_1)\text{ where }{X}_1\text{ is upper part of data}\\ \text{=45}\\ Q_3=Median(X_2),\text{ where }{X}_2\text{ is lower part of the data}\\ \text{=115}\\ IQR=115-45\\ =70\end{array}$

The five-number summary calculated as:

  $\begin{array}{l}=\{Minimum,Q_1,Median,Q_3,Maximum\}\\ =\{39,45,65,115,238\}\end{array}$

b.

The mean and standard deviation.

The mean is

The standard deviation is

Given:

The calculations are as follows:

S.No	$X$	$X^2$
1	39	1521
2	40	1600
3	43	1849
4	43	1849
5	44	1936
6	44	1936
7	45	2025
8	46	2116
9	46	2116
10	55	3025
11	62	3844
12	63	3969
13	65	4225
14	67	4489
15	79	6241
16	98	9604
17	105	11025
18	115	13225
19	115	13225
20	146	21316
21	176	30976
22	177	31329
23	181	32761
24	211	44521
25	238	56644
SUM	2343	307367

The Mean can be calculated as follows:

  $\begin{array}{l}\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}\\ =\frac{1}{25}\times 2343\\ =93.72\end{array}$

The Standard Deviation can be calculated as follows:

  $\begin{array}{l}s=\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}\\ =\sqrt{\frac{1}{24}\times \left({\displaystyle \sum _{i=1}^n{x}_i^2}-n{\overline{x}}^2\right)}\\ =\sqrt{\frac{1}{24}\times \left(307367-25\times {93.72}^2\right)}\\ =60.4776\end{array}$

c.

The mean and median and determine the relationship is true.

As the data is skewed to right, the relationship between mean and median is $Mean>Median$

The histogram for the data is

  

The data is skewed to right.

The computed mean is $93.72$

The computed median is $65$ .

  $\begin{array}{l}\text{Hence the relation between mean and median is:}\\ Mean>Median\end{array}$

d.

Whether the median and IQR or the mean and standard deviation is used to summarize the data.

The median and IQR are used to summarize the data.

Concept Used:

When there are outliers present in the data, median and IQR are used to summarize the data.

The data value that is more than $1.5\times IQR$ below $Q_1$ or above $Q_3$ is called outlier.

Explanation:

The presence of outliers in the data is identified using the IQR.

The calculated IQR is $555$ .

  $\begin{array}{c}\text{Lower Limt=}{Q}_1-1.5\times IQR\\ =45-1.5\times 70\\ =-60\\ \text{Upper Limt=}{Q}_3+1.5\times IQR\\ =115+1.5\times 70\\ =220\end{array}$

  $\left\{238\right\}\text{lie outside the limits and it is outliers}\text{.}$

In presence of outliers, the median and IQR are used to summarize the data.